Problem 116 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let the population of rabbits surviving at a time \(t\) be governed by the differential equation \(\frac{d p(t)}{d t}=\frac{1}{2} p(t)-200\). If initially \(p(0)=100\), then \(p(t)\) equals (A) \(400-300 e^{t / 2}\) (B) \(300-200 e^{-d 2}\) (C) \(600-500 e^{t / 2}\) (D) \(400-300 e^{-t / 2}\)

Step-by-Step Solution

Answer

The correct answer is (C) \(600 - 500e^{t/2}\).

Step 1: Identify the Type of Differential Equation

The given differential equation \( \frac{dp(t)}{dt} = \frac{1}{2}p(t) - 200 \) is a linear first-order differential equation of the form \( \frac{dy}{dt} + Py = Q \). Here \( P = -\frac{1}{2} \) and \( Q = -200 \).

Step 2: Find the Integrating Factor

To solve the differential equation, we find an integrating factor, \( \mu(t) \), which is given by \( e^{\int P \, dt} = e^{-\frac{1}{2}t} \).

Step 3: Solve the Differential Equation

Multiply the entire differential equation by the integrating factor: \[ e^{-\frac{1}{2}t} \cdot \frac{dp(t)}{dt} + e^{-\frac{1}{2}t} \cdot \frac{1}{2}p(t) = -200e^{-\frac{1}{2}t}. \]This simplifies to \[ \frac{d}{dt}(e^{-\frac{1}{2}t} \cdot p(t)) = -200e^{-\frac{1}{2}t}. \]

Step 4: Integrate to Find the General Solution

Integrate both sides of the equation:\[ \int \frac{d}{dt}(e^{-\frac{1}{2}t} \cdot p(t)) \, dt = \int -200e^{-\frac{1}{2}t} \, dt. \]The left side simplifies to \( e^{-\frac{1}{2}t} \, p(t) \), and the right side integrates to \[ -400e^{-\frac{1}{2}t} + C, \]where \( C \) is the constant of integration.

Step 5: Solve for the Particular Solution

The solution is now \[ e^{-\frac{1}{2}t} \, p(t) = -400e^{-\frac{1}{2}t} + C. \]Multiply through by \( e^{\frac{1}{2}t} \) to solve for \( p(t) \):\[ p(t) = -400 + Ce^{\frac{1}{2}t}. \]

Step 6: Apply Initial Condition

Use the initial condition \( p(0) = 100 \) to solve for \( C \):\[ 100 = -400 + C \cdot 1, \]which gives \( C = 500 \).

Step 7: Write the Final Solution

Substituting \( C = 500 \) back into the equation, we get:\[ p(t) = -400 + 500e^{t/2}. \]Rewriting gives:\[ p(t) = 500e^{t/2} - 400. \]This corresponds to option (C).

Key Concepts

First-order differential equations

A first-order differential equation involves a function and its first derivative. It's like having a story of how something changes without involving changes in changes (which would be second-order and so on).
In our rabbit population example, the function representing the number of rabbits at any time, noted as \(p(t)\), is connected to its rate of change, \(\frac{dp(t)}{dt}\). This setup keeps things simple, focusing only on how the population grows or shrinks, not the acceleration or deceleration of this change.
Such equations often appear as \(\frac{dy}{dt} + Py = Q\), where \(y\) changes with respect to time \(t\). Here, \(P\) and \(Q\) are usually constants or functions of \(t\).
First-order differential equations are crucial since they model processes like population growth, radioactive decay, cooling laws, and more.

Integrating factors

Integrating factors are a smart trick to simplify the solving of first-order linear differential equations. Think of them as a mathematical magic wand that transforms the equation into a form we can easily integrate.
In first-order linear equations, the goal is to make the equation look like the derivative of a product. For our equation, \(\frac{dp(t)}{dt} = \frac{1}{2}p(t) - 200\), we find an integrating factor \(\mu(t)\) which is \(e^{-\int P dt} = e^{-\frac{1}{2}t}\).
This factor allows us to rewrite our equation to reveal hidden simplicity: \(\frac{d}{dt}(\mu(t)p(t))\).
Once transformed, it’s much easier to integrate, giving us the pathway toward solving the differential equation for \(p(t)\).

Initial value problem

An initial value problem (IVP) involves not just a differential equation but also a point or 'starting value' that the solution must pass through. This starting point is essential in pinning down the exact solution from a family of possible solutions.
In our rabbit case, we know \(p(0) = 100\), meaning that at time \(t=0\), the rabbit population is 100.
This extra bit of information allows us to solve for any unknown constants in our general solution, ensuring our answer is unique and spot-on for the situation being modeled.
Thus, by combining the solution to the differential equation with the initial condition, we find a specific solution that satisfies both the rule of how things change and where they start from.